Evaluate
\[ \Large{\sum _{ n=1,3,5,7.... }^{ \infty }{ \frac { { e }^{ \left( 2n-1 \right) } }{ \int _{ 0 }^{ n+1 }{ \frac { { x }^{ e }{ e }^{ x } }{ \left( x+1 \right) ! } dx } } } }\]
Evaluate
\[ \Large{\sum _{ n=1,3,5,7.... }^{ \infty }{ \frac { { e }^{ \left( 2n-1 \right) } }{ \int _{ 0 }^{ n+1 }{ \frac { { x }^{ e }{ e }^{ x } }{ \left( x+1 \right) ! } dx } } } }\]
Note by
Lakshya Sinha
1 year, 8 months ago
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Top Newest@Jonas Katona – Lakshya Sinha · 1 year, 8 months ago
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I'm currently busy now. Ty using the infinite product of gamma(x+2) – Aditya Kumar · 1 year, 8 months ago
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What have you tried? What makes you think there's a closed form? – Pi Han Goh · 1 year, 8 months ago
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I don't know I just made this problem and asked my school teacher, she said it was pretty tough. So I asked for help. – Lakshya Sinha · 1 year, 8 months ago
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Not all series /integrals has a nice closed form. – Pi Han Goh · 1 year, 8 months ago
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try this – Lakshya Sinha · 1 year, 8 months ago
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See the report. Your problem is flawed. – Pi Han Goh · 1 year, 8 months ago
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Sorry – Lakshya Sinha · 1 year, 8 months ago
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@Tanishq Varshney,@Otto Bretscher,@Brian Charlesworth,@Michael Mendrin,@Aditya Kumar – Lakshya Sinha · 1 year, 8 months ago
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@Jake Lai – Lakshya Sinha · 1 year, 8 months ago
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Very unlikely this series has a closed form. Factorials/gamma functions in integrals, especially in denominators of integrals, are difficult to evaluate in general.
It is best to proceed in the direction of a motivating problem which applies integration in its solution, or to build a problem on the back of techniques already well-studied. – Jake Lai · 1 year, 8 months ago
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I made it accidently – Lakshya Sinha · 1 year, 8 months ago
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@Pi Han Goh – Lakshya Sinha · 1 year, 8 months ago
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